Defining on $R^3$, $V = \iiint_S dx \, dy \, dz $ as the volume of surface $S$, with $S$ closed, bounded and arc-connected. Which is the $S$ of minimal area, that contains $V$. I know it's a bit general, so maybe you could think of another restriction without losing too much generality
The link given by Brad is the isoperimetric inequality, which states that of all surfaces with area $A$, the one enclosing the most volume is a sphere. This also implies the answer to your question: of all surfaces enclosing a given volume $V$, the one with least surface area is a sphere.
To see this, fix a volume $V$ and let $S$ be any surface enclosing volume $V$. Let $S_0$ be a sphere having the same surface area as $S$. By the isoperimetric inequality, $S_0$ encloses more volume than $S$ (or the same amount). So if we scaled down $S_0$ to get a smaller sphere $S_1$ whose volume is $V$, it would have smaller surface area than $S_0$, and hence smaller surface area than $S$.
So we have shown that the sphere $S_1$ whose volume is $V$ has smaller surface area than any other surface $S$ enclosing volume $V$.
The relationship between statements like this is called duality.